Of sin 2 sin 2 indep of sin 2 thus

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: a) First, if B − A ∈ Z then for any α we have S B B cos 2π ( + α) 1 [cos(2πA+2πα)−cos(2πB+2πα)] =− = 2π 2π =A 1 [cos(2πA + 2πα) − cos(2πA + 2πα + 2π (B − A))] = 0 = 2π sin 2π ( + α) A ∈Z since cos has period 2π . On the other hand, suppose B − A ∈ Z. Then there exists an α, that is α = −A for which we get, using the same calculation, B sin 2π ( + α) A == 1 [cos(2πA + 2πα) − cos(2πA + 2πα + 2π (B − A))] = 2π 1 1 [cos 0 − cos 2π (B − A)] = [1 − cos 2π (B − A)] = 0 2π 2π ∈Z =1 (b) We have [ ×[ sin 2π ( + α) · sin 2π ( + β) = sin 2π ( + α) · sin 2π ( + β) = indep. of sin 2π ( + β) sin 2π ( + α) · = indep. of sin 2π ( + β) · Thus, using the result from part (a), we see that we have and only if [ ×[ sin 2π ( + α) · sin 2π ( + β) for any α β ∈ R. 10 sin 2π ( + α) − ∈ Z or =0 − ∈ Z if − (c) Since by assumption, [ ∈ Z or − ∈ Z, part (b) tells us that sin 2π ( + α) · sin 2π ( + β) ×[ for all α, β. (d) Since the small rectangles [×[ don’t overlap and their union is the big rectangle [×[, we have by the additivity of integration and by part (c) that [ ×[ sin 2π ( + α) · sin 2π ( + β) = =1 [ ×[ sin 2π ( + α) · sin 2π ( + β) =0 for all α β ∈ R. (e) Using part (d) together with part (b) again, tells us that E − ∈ Z or − ∈ Z. :) Evaluate...
View Full Document

This document was uploaded on 02/10/2014.

Ask a homework question - tutors are online